Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct edge {
  int u, v, next;
} edge[maxn * 2];
int head[maxn], cnt;
int n, m, q;
vector<int> person[maxn][20];
vector<int> city[maxn];
void addedge(int u, int v) {
  edge[cnt].u = u;
  edge[cnt].v = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
int deep[maxn], ances[maxn][20];
void update(vector<int>& a, vector<int>& b) {
  for (int i = 0; i < b.size(); i++) a.push_back(b[i]);
  sort(a.begin(), a.end());
  while (a.size() > 10) a.pop_back();
}
void dfs(int u) {
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (!deep[v]) {
      deep[v] = deep[u] + 1;
      ances[v][0] = u;
      update(person[v][0], city[v]);
      dfs(v);
    }
  }
}
void init() {
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i <= n; i++) {
      if (ances[i][j - 1] != -1) {
        update(person[i][j], person[i][j - 1]);
        update(person[i][j], person[ances[i][j - 1]][j - 1]);
        ances[i][j] = ances[ances[i][j - 1]][j - 1];
      }
    }
  }
}
vector<int> S;
int lca(int a, int b) {
  if (deep[a] < deep[b]) swap(a, b);
  int i, j;
  for (i = 0; (1 << i) <= deep[a]; i++)
    ;
  i--;
  for (j = i; j >= 0; j--) {
    if (deep[a] - (1 << j) >= deep[b]) {
      update(S, person[a][j]);
      a = ances[a][j];
    }
  }
  if (a == b) return a;
  for (j = i; j >= 0; j--) {
    if (ances[a][j] != -1 && ances[a][j] != ances[b][j]) {
      update(S, person[a][j]);
      update(S, person[b][j]);
      a = ances[a][j];
      b = ances[b][j];
    }
  }
  update(S, person[a][0]);
  update(S, person[b][0]);
  return ances[a][0];
}
int main() {
  memset(head, -1, sizeof(head));
  memset(ances, -1, sizeof(ances));
  cnt = 0;
  scanf("%d%d%d", &n, &m, &q);
  int u, v, a;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    scanf("%d", &c);
    city[c].push_back(i);
  }
  vector<int> emp;
  for (int i = 1; i <= m; i++) {
    update(city[i], emp);
  }
  deep[1] = 1;
  dfs(1);
  init();
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    S.clear();
    int lc = lca(u, v);
    update(S, city[lc]);
    int k = S.size() > a ? a : S.size();
    printf("%d ", k);
    for (int i = 0; i < k; i++) {
      printf("%d ", S[i]);
    }
    printf("\n");
  }
  return 0;
}